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[code]<?
require_once("db_connect.php");
$query = mysql_query("SELECT * FROM site");
$result = mysql_fetch_array($query);
$image_1 = $result['banner'];
$image_2 = imagecreatefrompng("$image_1");
imagepng(".$image_2.");
?>[/code]

That is my code. And this is the error I get on the page:

Warning: imagecreatefrompng(None Yet): failed to open stream: No such file or directory in /home/magneo/public_html/banner.php on line 6

Warning: imagepng(): supplied argument is not a valid Image resource in /home/magneo/public_html/banner.php on line 7

Could some one please tell me what I am doing wrong?
Link to comment
https://forums.phpfreaks.com/topic/5746-image-include-script/
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[!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]<?
require_once("db_connect.php");
$query = mysql_query("SELECT * FROM site");
$result = mysql_fetch_array($query);
$image_1 = $result['banner'];
$image_2 = imagecreatefrompng("$image_1");
imagepng("$image_2");
?>[/quote]

You put . before and afte rthe variable name inside a string so it would look like your trying to echo the .'s as part of the vpng to display therefore totaly corupting it..
Link to comment
https://forums.phpfreaks.com/topic/5746-image-include-script/#findComment-20494
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